Question

Find the values of x for which the function f(x) = x^3-3x ,is increasing.

Answer 1

The function is increasing if its derivative is positive.

$f'(x)=(x^3-3x)'=3x^2-3$

$3x^2-3>0$

$3x^2>3$

$x^2>1$

$|x|>1$

$x \in (- \infty ;-1)∪(1;+ \infty)$

Answer: $x \in (- \infty ;-1)∪(1;+ \infty)$